HW 8

nutrition <- read.csv("/Users/zen/Downloads/NutritionStudy.csv")
fish <- read.csv("/Users/zen/Downloads/FishGills3.csv")

Problem 1

We test whether the two alleles (R and X) are equally likely.

Hypotheses:

\(H_0: p_R = p_X\)

\(H_a: p_R \ne p_X\)

observed <- c(244, 192)
chisq.test(x = observed, p = c(0.5, 0.5))

## 
##  Chi-squared test for given probabilities
## 
## data:  observed
## X-squared = 6.2018, df = 1, p-value = 0.01276

p value:

0.01276

Conclusion:

Since the p value is less than 0.05, we reject H₀. There is enough evidence to say the two alleles (R and X) are not equally likely.

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Problem 2

We test if there is an association between vitamin use and sex.

Hypotheses:

\(H_0\): VitaminUse and Sex are independent

\(H_a\): VitaminUse and Sex are associated

table_data <- table(nutrition$VitaminUse, nutrition$Sex)
chisq.test(table_data)

## 
##  Pearson's Chi-squared test
## 
## data:  table_data
## X-squared = 11.071, df = 2, p-value = 0.003944

p value:

0.003944

Conclusion:

Since the p value is less than 0.05, we reject H₀. There is enough evidence to conclude that vitamin use and sex are associated.

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Problem 3

We test if mean gill rate differs by calcium level.

Hypotheses:

\(H_0: \mu_1 = \mu_2 = \mu_3\)

\(H_a\): At least one mean is different

anova_result <- aov(GillRate ~ Calcium, data = fish)
summary(anova_result)

##             Df Sum Sq Mean Sq F value Pr(>F)  
## Calcium      2   2037  1018.6   4.648 0.0121 *
## Residuals   87  19064   219.1                 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

p value:

0.0121

Conclusion:

Since the p value is less than 0.05, we reject H₀. There is enough evidence to say that mean gill rate differs across calcium levels.